14 Ellipse

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ELLIPSE

65. The equation2 2x y

110 a 4 a

+ =- -

represents an ellipse if-

•(1) a < 4 (2) a > 4 (3) 4 < a < 10 (4) a > 10

65. lehdj.k 2 2x y

110 a 4 a

+ =- -

nh?kZo`Ùk dks iznf'k Zr djsxk fn -

(1) a < 4 (2) a > 4 (3) 4 < a < 10 (4) a > 10

65. Ans. (1)

10 – a > 0 and 4 – a > 0

Þ a < 470. Eccentricity of an ellipse whose latus rectum is half of its minor axis is -

(1)

1

2 (2)2

3 •(3)3

2 (4)1

2

70. nh?kZo`Ùk dh mRdsUærk ftldk ukfHkyEc blds y?kq v{k dk vk/kk gS gks xh -

(1)1

2(2)

2

3(3)

3

2(4)

1

2

70. Ans. (3)

( )22b 1

2ba 2

=

2b = a4b2 = a2

4a2(1 – e2) = a2

3e

2=

73. Tangents are drawn from points on the circle x2 + y2 = 49 to the ellipse2 2x y

125 24

+ = , then angle

between the tangents is-

(1)4p •(2)

2p (3)

3p (4)

8p

73. o`Ùk x2 + y2 = 49 ij fLFkr fcUnqvks a ls nh?kZo`Ùk 2 2x y

125 24

+ = ij Li'kZ js [kk;s a khaph tkrh gS rks Li'kZ js [kkvks a ds e

dks .k gks xk -

(1)4

p(2)

2

p(3)

3

p(4)

8

p

73. Ans. (2)

x2 + y2 = 49 is director circle of2 2

x y 125 24

+ =

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76. If3

p and a are eccentric angles of the ends of a focal chord of ellipse

2 2

2 2

x y1

a b+ = , a > b and if its

eccentricity is1

2, then tana can be-

(1)1

3•(2) 3- (3)

2

3(4)

2

3-

76. ;fn 3

p rFkk a nh?kZo`Ùk

2 2

2 2

x y1

a b+ = , a > b dh ukHkh thok ds vfUre fljks a ds mRds Uæ dks.k gS rFkk fn bldh

mRds Uærk 1

2 gS rks tana dk eku fuEu gks ldrk gS -

(1)

1

3 (2) 3- (3)

2

3 (4)

2

3-

76. Ans.(2)

e 1 1tan tan

6 2 e 1 3

p a -= = -

+

1tan

2 3

a= -

2

2tan

2tan 31 tan

2

a

a = = -a-

83. The eccentric angles of extremities of latus rectum of ellipse2 2x y

125 16

+ = are given by -

(1) 13

tan5

- æ ö±ç ÷è ø

(2) 112

tan5

- æ ö±ç ÷è ø

•(3) 14

tan3

- æ ö±ç ÷è ø

(4) 125

tan12

- æ ö±ç ÷è ø

83. nh?kZo`Ùk 2 2x y

1

25 16

+ = ds ukfHkyEc ds fljks a ds mRds Uæ dks .k fuEu kjk fn;k tkrs gS -

(1) 13

tan5

- æ ö±ç ÷è ø

(2) 112

tan5

- æ ö±ç ÷è ø

(3) 14

tan3

- æ ö±ç ÷è ø

(4) 125

tan12

- æ ö±ç ÷è ø

83. Ans. (3)

5 cosq = ±ae16

4sin5

q = ±

cosq = ±3

5

4sin

5q = ±

4tan3

æ öq = ± ç ÷è ø

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Q U E S T I O N

B A

N K

2 0 1 4

88. Statement-I : In an ellipse the distance between foci is always less than the sum of focal distance of any point on it.

Statement-II : If e is eccentricity of the ellipse, then 0 < e < 1.

•(1) Statement-I is true, Statement-II is true; statement-II is a correct explanation for Statement-I.

(2) Statement-I is true, Statement-II is true; statement-II is not a correct explanation for Statement-I.

(3) Statement-I is true, Statement-II is false.(4) Statement-I is false, Statement-II is true.

88. dFku -I : nh?kZo`Ùk es a ukfHk;ks a ds e nwjh lnS o blds fdlh fcUnq dh ukHkh nwfj;ks a ds ks xQy ls de gks xhA

dFku-II : ;fn e, nh?kZo`Ùk dh mRds Uærk gS rks 0 < e < 1 gksxkA

(1) dFku -I lR gS dFku -II lR gS dFku -II dFku -I dh g O;k k gSA(2) dFku -I lR gS dFku -II lR gS dFku -II dFku -I dh lgh O;k k ug a gSA(3) dFku -I lR gS dFku -II vlR gSA(4) dFku -I vlR gS dFku -II lR gSA

88. Ans. (1)

sincec

e 1a

= < Þ c < a

5. If circle whose diameter is major axis of ellipse2 2

2 2

x y1

a b+ = (a > b) meets minor axis at point P &

orthocentre of DPF1F2 lies on ellipse where F1 & F2 are focii of ellipse, then square of eccentricityof ellipse is-

;fn nh?kZo`Ùk 2 2

2 2

x y1

a b+ = (a > b) dh nh?kZv{k dks O;kl ekudj khapk x;k o`Ùk y?kqv{k dks fcUnq P ij feyrk gS rFkk

f=Hkqt PF1F2 dk yEcds Unz nh?kZo`Ùk ij fLFkr gS tgk¡ F1 rFkk F2 nh?kZo`Ùk dh ukfHk;k¡ gS rks nh?kZo`Ùk dh mRds Unzrk dk oxZ gksxk -•(A) 2 sin18º (B) 2 sin15º (C) sin45º (D) sin60º

5. Ans. (A)

2 1F H PFm .m 1= - P(0,a)

F2 F1Þ b a

1ae ae

æ ö- = -ç ÷è ø

Þ 2b e

a=

Þ 1 – e2 = e4 Þ e4 + e2 – 1 = 0

Þ 21 5

e 2sin182

- += = °

6. An ellipse has foci at (3,4) and (7,5) and has x-axis as its tangent. Its eccentricity is-

fdlh nh?kZo`Ùk dh ukfHk;k¡ (3,4) rFkk (7,5) ij fLFkr gS rFkk x-v{k bldh Li'kZ js [kk gks rks nh?kZo`Ùk dh mRds Unzrk gks xh -

(A)20

97

(B)10

97

(C)5

97

•(D)17

976. Ans. (D)

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Q U E S T I O N

B A

N K

2 0 1 4

p1

p1

(3,4)(7,5)

x-axis

p1p2 = b2 Þ b2 = 20

2ae 17=

b2 = a2 – a2e2 Þ 2 97a

4=

17c

97=

2. If P1, P2, P3 are the points on ellipse 3x2 + y2 – 12 = 0 and P1', P2', P3' are their corresponding points

on the auxillary circle, then the area of triangle P1' P2' P3' is l times the area of triangle P1P2P3, thenl2 is

;fn P1, P2, P3 nh?kZo`Ùk 3x2 + y2 – 12 = 0 ij fLFkr fcUnq gS rFkk P1', P2', P3' lgk;d òÙk ij fLFkr buds la xr fcUnq gS rks f=Hkqt P1' P2' P3' dk ks=Qy f=Hkqt P1P2P3 ds ks =Qy dk l xquk gks rks l2 dk eku gksxk

Ans. 3

2. Ans. 3

Let ( )1 1 1P 2 cos 2 3 sinq q p2 p1

p3

p21

p11

p31

( )11 1 1P 2 3 cos' ', 2 3 sinq q

Area

1 1

1 1 11 2 3 2 2

3 3

2 3 cos 2 3 sin 11

P P P 2 3 cos 2 3 sin 12

2 3 cos 2 3 sin 1

q q

= q q

q q

1 1

1 2 3 2 2

3 3

2cos 2 3 sin 11

P P P 2cos 2 3 sin 12

2 cos 2 3 sin 1

q q

= q q

q q

22 3.2 3 3 32.2 3

l = = Þ l =

18. If tangents drawn from point (1, 2) to ellipse x2 + 2y2 = 1 cuts x-axis at points A(x1, 0) & B(x

2, 0),

then H.M. of x1 & x

2 is -

(1) –9 •(2) 9 (3) –7 (4) 7

18. ;fn fcUnq (1, 2) ls nh?kZo`Ùk x2 + 2y2 = 1 ij khaph xbZ Li'kZ js [kk;sa x- v{k dks fcUnq A(x1, 0) rFkk B(x2, 0) ij dkVrh gS rks x1 rFkk x2 dk gjkRed ek gks xk (1) –9 (2) 9 (3) –7 (4) 7

18. Ans. (2)

Pair of tangents SS1 = T2

8(x2 + 2y2 – 1) = (x + 4y – 1)2

put y = 0 Þ 7x2 + 2x – 9 = 0

1 2

1 2

2x xH.M. 9

x x= =

+

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Q U E S T I O N

B A

N K

2 0 1 4

23. Tangents are drawn from every point on the line x + 9y = 4 to the ellipse2 2x 9y

14 4

+ = , then the

corresponding chords of contact always pass through (a,b), then value of (a + b) is-(1) 0 •(2) 2 (3) 5 (4) 7

23. js [kk x + 9y = 4 ds lHkh fcUnqvks a ls nh?kZo`Ùk 2 2x 9y

14 4

+ = ij Li'kZ js [kk;sa khpha tkrh gS rc laxr Li'kZ thok lnS o

(a,b) ls xqtjrh gS rks (a + b) dk eku gksxk (1) 0 (2) 2 (3) 5 (4) 7

23. Ans. (2)

x + 9y = 4P(4 – 9k, k) as a point in given lineEquation of chord of cantact

(4 9k)x 9ky

14 4

-

+ =(4 – 9k)(x) + 9ky = 4

4(x – 1) + 9k(y – x) = 0

x – 1 = 0

y – x = 0

So, point will (1, 1)

25. Normal at variable point P on ellipse 2x2+y2= 1 meets the coordinate axes at Q & R, then eccentricity

of locus of mid point of QR will be-

•(1)1

2 (2)1

3 (3)1

2 (4) 2

25. fdlh nh?kZo`Ùk 2x2+y2= 1 ds pj fcUnq P ij khapk x;k vfHkyEc funs'khZ v{kksa dks Q rFkk R ij feyrk gS rc QR ds e fcUnq ds fcUnqiFk dh mRds Unzrk gks xh -

(1)1

2(2)

1

3(3)

1

2(4) 2

25. Ans. (1)

Locus of mid point of QR will be a similar ellipse.So eccentricity will be same as thatif original ellipse

2 2x y1

1 2+ =

1 1e 1

2 2= - =

26. If center of ellipse( ) ( )

2 2x 3y 5 3x y 5

410 20

+ - - -+ = is (a,b), then 2a + b will be -

(1) 1 (2) 2 (3) 3 •(4) 5

26. ;fn nh?kZo`Ùk ( ) ( )

2 2x 3y 5 3x y 5

410 20

+ - - -

+ = dk ds Unz (a,b) gks rks 2a + b dk eku gksxk (1) 1 (2) 2 (3) 3 (4) 5

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Q U E S T I O N

B A

N K

2 0 1 4

26. Ans. (4)

x + 3y – 5 = 03x – y – 5 = 0are principal axes of ellipseSo their point of intersection will be center

i.e. (2, 1)78. F

1 & F

2 are foci of an ellipse and B is on extrimity of minor axis. If F

1 B F

2 is a right angle isosceles

triangle then eccentricity of ellipse is-

(1)1

2•(2)

1

2(3)

3

2(4)

1

2 2

78. F1 rFkk F

2 nh?kZo`Ùk dh ukfHk;k¡ rFkk B y?kqv{k ds fljs ij gSA fn F

1 B F

2 ,d ledks .k lef}ckgq f=Hkqt gks rks nh?kZo`Ùk

dh mRds Uærk gks xh -

(1) 12 (2)1

2 (3) 32(4)

1

2 2

78. Ans. (2)

F1F

2 = 2a 2ae=

(0,b)

F2 F1

Þ1

e2

=

82. If curves2 2x y

149 25

+ = and2 2

2

x y1

15 k- = are orthogonal to each other, then possible values of k are-

(1) ±1 (2) ±2 •(3) ±3 (4) ±4

82. ;fn oØ 2 2x y

149 25

+ = rFkk 2 2

2

x y1

15 k- = yEcdks .kh gks rks k ds lEHko eku gks axs -

(1) ±1 (2) ±2 (3) ±3 (4) ±4

82. Ans. (3)

1

2 2

2 21

x y1

a b+ = &

2 2

2 22 2

x y1

a b- =

are orthogonal then2 2 2 21 1 2 2a b a b- = + Þ 49 – 25 = 15 + k

2 Þ k = ±3

88. e1 is eccentricity of curve2 2x y

149 24

+ = and e2 is eccentricity of curve2 2x y

124 49

+ = .

Statement-I : Value of e1 is equal to that of e2.

Statement-II : Eccentricity of ellipse decreases as the ratio of length of minor axis and major axis

increases.

(1) Statement-I is true, Statement-II is true; statement-II is a correct explanation for Statement-I.

•(2) Statement-I is true, Statement-II is true; statement-II is not a correct explanation for Statement-I.

(3) Statement-I is true, Statement-II is false.

(4) Statement-I is false, Statement-II is true.

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88. e1, oØ 2 2x y

149 24

+ = dh mRds Uærk rFkk e2, oØ 2 2x y

124 49

+ = dh mRds Uærk gSA

dFku-I : e1 dk eku e2 ds cjkcj gSA

dFku

-II : nh?kZo`Ùk dh mRds Uærk y?kqv{k rFkk nh?kZv{k dh yEckbZ ds vuqikr es a o`f¼ ds lkFk kVrh gSA(1) dFku -I lR gS dFku -II lR gS dFku -II dFku -I dh g O;k k gSA

(2) dFku -I lR gS dFku -II lR gS dFku -II dFku -I dh lgh O;k k ug a gSA

(3) dFku -I lR gS dFku -II vlR gSA

(4) dFku -I vlR gS dFku -II lR gSA88. Ans. (2)

2

1 2

be e ,e 1

a

æ ö= = -ç ÷è ø

Paragraph for Question 10 to 11

Consider the ellipse whose equation is2 2

2

x y1

16 b+ = , then

ekuk nh?kZo`Ùk 2 2

2

x y1

16 b+ = gSA

10. If PQ is a variable chord of the above ellipse and PQ subtends an angle 90º at the origin, also given

2 2

1 1 25

OP OQ 144+ = , then b2

is-

(A) 4 •(B) 9 (C) 25 (D) None of these

;fn PQ mijks ä nh?kZo` Ùk dh d pj thok gS rFkk PQ, ew yfcUnq ij 90º dks .k vUrfjr djrh gS rFkk 2 21 1 25

OP OQ 144+ =

gks rks b2 dk eku gksxk (A) 4 •(B) 9 (C) 25 (D) bues a ls dks bZ ugha

11. A rectangle whose sides are parallel to co-ordinate axes, circumscribes the ellipse with b2 obtained

in above problem. An ellipse through (6,0) circumscribes the above rectangle, one of its corner

point being (4,3). Then the equation of the second ellipse is-

(A)2 2x y

136 81

+ = (B)2 2x y

136 16

+ = •(C)2 2x 5y

136 81

+ = (D) None of these

,d vk;r ftldh Hkqtk;sa funs Z'khZ v{kksa ds lekUrj gS mijks Dr iz'u es a izkIr b2 okys nh?kZo`Ùk ds ifjxr gSA (6,0) ls xqtjus okyk d vU nh?kZo`Ùk bl vk;r ds ifjxr gS ftldk d dksuk (4,3) ij fLFkr gSA rc nwljs nh?kZo`Ùk dk lehdj.k gks xk

(A)2 2x y

1

36 81

+ = (B)2 2x y

1

36 16

+ = •(C)2 2x 5y

1

36 81

+ = (D) None of these

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Q U E S T I O N

B A

N K

2 0 1 4

Paragraph for Question 10 to 11

10. Ans. (B)

P

q

Q

90º

O

Let OP = r

\ P(r cosq, rsinq) lies on ellipse2 2

2

x y1

16 b+ = .

Þ 2 2

2 2

1 cos sin

OP 16 b

q q= + ...(1)

for the point Q replace q ® 90 + q

\ 2 2

2 2

1 sin cos

OQ 16 b

q q= + ...(2)

\ (1) + (2) Þ 225 1 1

144 16 b= +

Þ b2 = 911. Ans. (C)

(4,3)

(6,0)

Equation of second ellipse is

2 2

2x y 136 B+ = but (4,3) lies on it

Þ 216 9

136 B

+ = Þ 229 5 81

BB 9 5

= Þ =

\ Equation of second ellipse2 2x 5y

136 81

+ = .

2. An ellipse with focii (1,4) and (a,b) touches x-axis at (5,0). Then value of (a – b) is

,d nh?k Zo`Ùk ftldh ukfHk;k¡ (1,4) rFkk (a,b) gS x-v{k dks fcUnq (5,0) ij Li'kZ djrk gSA rc (a – b) dk eku gksxk

Ans. 5

2. Ans. 5

(1,4), (a,b) and reflection of (a,b) on x-axis and collinear

1 4 1

5 0 1 0 5

1

= Þ a - b =

a -b

12. An ellipse has its focii as S1(1,–2) and S

2(–3,4). If the foot of perpendicular dropped from S

2 on a

tangent to the ellipse is (1,6), then length of minor axis of ellipse will be -

(1) 2 units (2) 4 units •(3) 8 units (4) 16 units

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Q U E S T I O N

B A

N K

2 0 1 4

12. ,d nh?kZo`Ùk dh ukfHk;k¡ S1(1,–2) rFkk S2(–3,4) gSA fn nh?kZo`Ùk dh fdlh Li'kZ js [kk ij S2 ls Mkys x;s yEc dk ikn (1,6) gks rks nh?kZo`Ùk ds y?kqv{k dh yEckbZ gks xh (1) 2 bdkbZ (2) 4 bdkbZ (3) 8 bdkbZ (4) 16 bdkbZ

12. Ans. (3)

(1,6)

s1 s2

p1 p2

p1p2 = b2 (refer highlights)

2 22p 4 2 2 5= + =

slope of tangent : –2Equation of tangent : y – 6 = –2(x – 1)Þ 2x + y – 8 = 0

\ p1=

8

5

p1p

2 = 16 = b2 Þ b = 4

\ 2b = 8

13. Area of quadrilateral formed by joining focii of ellipses2 2x y

116 12

+ = and2 2x y

112 16

+ = is -

(in sq. units)

(1) 2 (2) 4 (3) 6 •(4) 8

13. nh?kZo`Ùk 2 2x y

116 12

+ = rFkk 2 2x y

112 16

+ = dh ukfHk;ks a dks feykus ij fufeZr prqHkq Zt dk ks =Qy gks xk -

( oxZ bdkbZ es a)

(1) 2 (2) 4 (3) 6 (4) 813. Ans. (4)

2 2x y1

16 12+ = Þ focii (±2, 0)

2 2x y1

12 16+ = Þ focii (0, ±2)

A =1

4 2 2 82

´ ´ ´ =

10. The difference between maximum and minimum values of 2x + y, where x and y satisfy the equation9x2 + 4y2 = 36 is-(1) 5 (2) 6 (3) 8 •(4) 10

10. 2x + y ds vf/kdre rFkk U;wure ekuks a ds e vUrj tgk¡ x rFkk y lehdj.k 9x2 + 4y2 = 36 dks lUrq"V djrs gS gksxk -(1) 5 (2) 6 (3) 8 (4) 10

10. Ans. (4)

2 2x y1

4 9+ = Þ x = 2cosq; y = 3sinq

\ E = 2(2cosq) + 3sinq = 4cosq + 3sinq

E Î [–5,5]

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Q U E S T I O N

B A

N K

2 0 1 4

15. Minimum distance between the line x + y = 7 and the ellipse x2 + 2y2 = 6 is-

(1) 2 •(2) 2 2 (3) 3 2 (4) 4 215. js [kk x + y = 7 rFkk nh?kZo`Ùk x2 + 2y2 = 6 ds e U;wure nwjh gks xh -

(1) 2 (2) 2 2 (3) 3 2 (4) 4 2

15. Ans. (2)

Ö q Ö q6cos , 3sinP

x6

2y3

2

+ =1

x+y=7

Slope of tangent :1

cot 1

2

- q = - Þ cot 2q =

\ 2 1

P 6 , 3 P(2,1)3 3

æ ö´ ´ Þç ÷ç ÷

è ø

Distance2 1 7

2 22

+ -æ ö= =ç ÷

è ø

16. a and b are eccentric angles of the ends of a focal chord of ellipse2 2x y

116 12

+ = . Value of

tan tan2 2

tan2

a bæ ö æ ö+ç ÷ ç ÷è ø è øa + bæ ö

ç ÷è ø

can be -

(1)4

3- (2)

1

3- •(3)

4

3(4)

1

3

16. ;fn a rFkk b nh?kZo`Ùk 2 2x y

116 12

+ = dh ukHkh thok ds fljks a ds mRØs Uæ dks .k gks rks tan tan

2 2

tan 2

a bæ ö æ ö+ç ÷ ç ÷è ø è ø

a + bæ ö

ç ÷è ø

dk eku gks

ldrk gS -

(1)4

3- (2)

1

3- (3)

4

3(4)

1

3

16. Ans. (3)

tan tan2 2 1 tan tan

2 2tan

2

a b+ a b

= -a + bæ ö

ç ÷è ø

for focal chord with eccentric points (a) & (b)

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Q U E S T I O N

B A

N K

2 0 1 4

e 1tan .tan

2 2 e 1

a b -=

+ or

e 1

e 1

+-

Þ 1

tan .tan2 2 3

a b= - or –3

\ value4

3= or 4.

18. A square ABCD is inscribed inside the ellipse2

2x y 13

+ = . Area of quadrilateral formed by tangents

at points A,B,C & D will be-(in sq. units)(1) 2 (2) 4 •(3) 8 (4) 16

18. nh?kZo`Ùk 2

2x y 1

3

+ = ds vUrxZr oxZ ABCD gSA fcUnqvks a A,B,C rFkk D ij khaph xbZ Li'kZ js [kkvks a kjk fufeZr

prqHkqZt dk ks=Qy gksxk ( oxZ bdkbZ es a) -(1) 2 (2) 4 (3) 8 (4) 16

18. Ans. (3)

D

C B

0

A( 3cos ,sin )Ö qq

For square 3 cos sinq = q

Þ tan 3q =

Þ 3

pq = (for A)

\ Equation of tangentx y 3

1 x 3y 2 322 3

+ = Þ + =´

Area of DOAB =1 2 3

2 3 2

2 3

´ ´ =

\ Ar. (WABCD)= 4 × 2 = 8

20. Product of slopes of common tangents to the curves 2y 16 3x= and2 2x y

12 4

+ = is -

•(1) –4 (2) –8 (3) –24 (4) –64

20. oØksa 2y 16 3x= rFkk 2 2x y

12 4

+ = dh mHk;fu"B Li'kZ js[kkvksa dh izo.krkvks a dk xq.kuQy gks xk -

(1) –4 (2) –8 (3) –24 (4) –6420. Ans. (1)

4 3y mx

m= +

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Q U E S T I O N

B A

N K

2 0 1 4

2y mx 2m 4= ± +

\ 24 3

2m 4m

± + =

Þ 2(m2

+ 2)m

2

= 48Þ m4 + 2m2 – 24 = 0 Þ m2 = –6; m2 = 4\ m = ±2 Þ m1m2 = –4

26. For DABC, if B is origin and coordinates of C are (8,0) and vertex A moves such that

B C3tan tan 1

2 2= . Length of latus rectum of conic represented by locus of A will be-

•(1) 12 (2) 15 (3) 16 (4) 18

26. f=Hkq t ABC ds fy fn B ewyfcUnq rFkk C ds funs Z 'kka d (8,0) ,oa 'kh"kZ A bl iz dkj xfr djrk gS fd B C

3tan tan 12 2

=

gSA A ds fcUnqiFk kjk iznf'kZr 'kkado ds ukfHkyEc dh yEckbZ gS -(1) 12 (2) 15 (3) 16 (4) 18

26. Ans. (1)

( ) ( )( )

( ) ( )( )

s a s c s a s bb c 1 1tan .tan .

2 2 3 s s b s s c 3

- - - -= Þ =

- -

Þ s a 1

b c 2as 3

-= Þ + = A

B(0) C(8,0)

Þ b + c = 16.\ A lies on an ellipse.

\ 2ae = 8, 2a = 16

Þ 2

21 be 1 b 482 64

= = - Þ =

\ LLR =22b 2 48

12a 8

´= =

7. If ax + by = 1 be the common tangent to the curves2 2x y

116

+ =l

, (y – 2)2 = x – 4 and (x – 3)2 + y2

= 1, then (8a + 3b) is equal to (where l is positive real number)

;fn ax + by = 1, oØksa 2 2x y

116

+ =l

, (y – 2)2 = x – 4 rFkk (x – 3)2 + y2 = 1 dh mHk;fu"B Li'kZ js [kk gks

rks (8a + 3b) dk eku gks xk ( tgk¡ l /kukRed okLrfod la k gS)Ans. 2

7. Ans. 2

Clearly, the line x = 4 is a common tangent for the curves. (y – 2)2 = x – 4 and (x – 3)2 + y2=1

and the line x = 4 also touches the given ellipse for all values of l Î R+.

\ equation of common tangent is1

x 0.y 14

+ =

Þ 1a , b 0 8a 3b 24

= = Þ + = .

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Q U E S T I O N

B A

N K

2 0 1 4

10. The difference between maximum and minimum values of 2x + y, where x and y satisfy the equation9x2 + 4y2 = 36 is-(1) 5 (2) 6 (3) 8 •(4) 10

10. 2x + y ds vf/kdre rFkk U;wure ekuksa ds e vUrj tgk¡ x rFkk y lehdj.k 9x2 + 4y2 = 36 dks lUrq"V djrs gS gks xk -

(1) 5 (2) 6 (3) 8 (4) 1010. Ans. (4)

2 2x y1

4 9+ = Þ x = 2cosq; y = 3sinq

\ E = 2(2cosq) + 3sinq = 4cosq + 3sinqE Î [–5,5]

15. Minimum distance between the line x + y = 7 and the ellipse x2 + 2y2 = 6 is-

(1) 2 •(2) 2 2 (3) 3 2 (4) 4 215. js [kk x + y = 7 rFkk nh?kZo`Ùk x2 + 2y2 = 6 ds e U;wure nwjh gks xh -

(1) 2 (2) 2 2 (3) 3 2 (4) 4 215. Ans. (2)

Ö q Ö q6cos , 3sinP

x6

2y3

2

+ =1

x+y=7

Slope of tangent :1

cot 12

- q = -

Þ cot 2q =

\ 2 1

P 6 , 3 P(2,1)3 3

æ ö´ ´ Þç ÷ç ÷

è ø

Distance2 1 7

2 2

2

+ -æ ö= =ç ÷

è ø

16. a and b are eccentric angles of the ends of a focal chord of ellipse2 2x y

116 12

+ = . Value of

tan tan2 2

tan2


a + bæ öç ÷è ø

can be-

(1)

4

3- (2)

1

3- •(3)

4

3 (4)

1

3

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Q U E S T I O N

B A

N K

2 0 1 4

16. ;fn a rFkk b nh?kZo`Ùk 2 2x y

116 12

+ = dh ukHkh thok ds fljks a ds mRØs Uæ dks .k gks rks tan tan

2 2

tan2


a + bæ öç ÷è ø

dk eku gks

ldrk gS -

(1)4

3- (2)

1

3- (3)

4

3(4)

1

3

16. Ans. (3)

tan tan2 2 1 tan tan

2 2tan

2

a b+ a b

= -a + bæ ö

ç ÷è ø

for focal chord with eccentric points (a) & (b)

e 1tan .tan

2 2 e 1

a b -=

+ or

e 1

e 1

+-

Þ 1

tan .tan2 2 3

a b= - or –3

\ value4

3= or 4.

18. A square ABCD is inscribed inside the ellipse2

2x y 1

3

+ = . Area of quadrilateral formed by tangents

at points A,B,C & D will be (in sq. units) -(1) 2 (2) 4 •(3) 8 (4) 16

18. nh?kZo`Ùk 2

2x y 13

+ = ds vUrxZr oxZ ABCD gSA fcUnqvks a A,B,C rFkk D ij khaph xbZ Li'kZ js [kkvks a kjk fufeZr

prqHkqZt dk ks=Qy gksxk ( oxZ bdkbZ es a) -(1) 2 (2) 4 (3) 8 (4) 16

18. Ans. (3)

D

C B

0

A( 3cos ,sin )Ö qq

For square 3 cos sinq = q

Þ tan 3q =

Þ 3

pq = (for A)

\ Equation of tangentx y 3 1 x 3y 2 3

22 3+ = Þ + =

´

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Q U E S T I O N

B A

N K

2 0 1 4

Area of DOAB =1 2 3

2 3 22 3

´ ´ =

\ Ar. (WABCD)= 4 × 2 = 8

20. Product of slopes of common tangents to the curves 2y 16 3x= and

2 2x y12 4+ = is-

•(1) –4 (2) –8 (3) –24 (4) –64

20. oØksa 2y 16 3x= rFkk 2 2x y

12 4

+ = dh mHk;fu"B Li'kZ js[kkvksa dh izo.krkvks a dk xq.kuQy gks xk -

(1) –4 (2) –8 (3) –24 (4) –6420. Ans. (1)

4 3y mx

m= +

2y mx 2m 4= ± +

\ 24 3

2m 4m

± + =

Þ 2(m2 + 2)m2 = 48Þ m4 + 2m2 – 24 = 0 Þ m2 = –6; m2 = 4\ m = ±2 Þ m

1m

2 = –4

26. For DABC, if B is origin and coordinates of C are (8,0) and vertex A moves such that

B C3tan tan 1

2 2

= . Length of latus rectum of conic represented by locus of A will be-

•(1) 12 (2) 15 (3) 16 (4) 18

26. f=Hkq t ABC ds fy fn B ewyfcUnq rFkk C ds funs Z 'kka d (8,0) ,oa 'kh"kZ A bl iz dkj xfr djrk gS fd B C

3tan tan 12 2

=

gSA A ds fcUnqiFk kjk iznf'kZr 'kkado ds ukfHkyEc dh yEckbZ gS -(1) 12 (2) 15 (3) 16 (4) 18

26. Ans. (1)

( ) ( )( )

( ) ( )( )

s a s c s a s bb c 1 1tan .tan .

2 2 3 s s b s s c 3

- - - -= Þ =

- -

Þ s a 1

b c 2as 3

-= Þ + =

A

B(0) C(8,0)Þ b + c = 16.\ A lies on an ellipse.\ 2ae = 8, 2a = 16

Þ 2

21 be 1 b 482 64

= = - Þ =

\ LLR

=22b 2 48

12a 8

´= =

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Q U E S T I O N

B A

N K

2 0 1 4

4. Given that M = {(x,y)| x2 + 2y2 = 3} and N = {(x,y)|y = mx + c}. If M Ç N ¹ f for all m Î R, thenrange of 'c' is-

•(A)6 6

,2 2

é ù-ê úë û

(B)6 6

,2 2

æ ö-ç ÷ç ÷

è ø(C)

6 6,

2 2

æ ù-ç úç

è û(D)

2 3 2 3,

3 3

é ù-ê úë û

ekuk M = {(x,y)| x2 + 2y2 = 3} rFkk N = {(x,y)|y = mx + c} gSA fn lHkh m Î R ds fy;s M Ç N ¹ f gks rks'c' dk ifjlj gksxk

•(A)6 6

,2 2

é ù-ê úë û

(B)6 6

,2 2

æ ö-ç ÷ç ÷

è ø(C)

6 6,

2 2

æ ù-ç úç

è û(D)

2 3 2 3,

3 3

é ù-ê úë û

4. Ans. (A)

{ }2 2M (x, y) | x 2y 3= + =

and N {(x,y) | y mx c}= = +

If M Ç N ¹ f " m Î R, then the point (0, c) lies within the ellipse2 2x y 1

3 3 / 2+ = ,

therefore22c

13

£ Þ6 6

C2 2

-£ £

2. Suppose F1 & F

2 are the foci of the ellipse 4x2 + 9y2 = 36. P is point on the ellipse such that

|PF1| : |PF

2| = 2 : 1, then area of DPF

1F

2 is

2. ekuk F1 rFkk F2, nh?kZo`Ùk 4x2 + 9y2 = 36 dh ukfHk;k¡ gSA fcUnq P, nh?kZo`Ùk ij bl izdkj gS |PF1| : |PF2| = 2 : 1 gks

rks f=Hkqt PF1F2 dk ks =Qy gks xk Ans. 004

2. Ans. 004

F1 F2

P

2 2x y1

9 4+ =

Þ a = 3, b = 2

Þ 5

e3

=

PF1 = 2l, PF2 = l2l + l = 6 Þ l = 2PF1 = 4, PF2 = 2

1 2

5F F 2.3. 2 5

3= =

PF1

2 + PF2

2 = F1F

22 Þ ÐF

1PF

2 = 90º

Þ area1

.4.2 42

= = PF2

| = 2 : 1, then area of DPF1

F2

is

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Q U E S T I O N

B A

N K

2 0 1 4

1. The equation of locus of circumcentre of the triangle formed by any tangent to the ellipse2 2x y

14 9

+ =

in first quadrant and coordinates axes, is -

•(A ) 9x2 + 4y2 = 4x2y2 (B) 4x2 – 9y2 = 4x2y2

(C) 4x2 + 9y2 = 4x2y2 (D) 9x2 – 4y2 = 4x2y2

izFke prqFkk±'k es a nh?kZo`Ùk 2 2x y

14 9

+ = dh fdlh Li'kZ js [kk rFkk funs Z'kh v{kksa kjk fufeZr f=Hkqt ds ifjds Uæ ds fcUnqiFk

dk lehdj.k gks xk -(A) 9x2 + 4y2 = 4x2y2 (B) 4x2 – 9y2 = 4x2y2

(C) 4x2 + 9y2 = 4x2y2 (D) 9x2 – 4y2 = 4x2y2

1. Ans. (A)

\ Tangent at 'q'

x ycos sin 1

2 3q + q =

Circumcentre is mid point of hypotenuse

1h

cos=

q

3k

2sin=

q

\ 2 21 9

1h 4k

+ =

Þ 9x2 + 4y2 = 4x2y2

9. If there exists a circle 'C' whose one diameter is the join of focii of ellipse2 2

2 2

x yE : 1

a b+ = (a > b) and

other diameter is the minor axis of E, then-

(A) eccentricity of E is1

2

•(B) eccentricity of E is 12

(C) If E is2 2

2

x y1

4 b+ = then director circle of circle 'C' is 2 2x y 8+ =

•(D) If E is2 2

2

x y1

4 b+ = then director circle of circle 'C' is 2 2x y 4+ =

;fn d o`Ùk 'C' fo|eku gS ftldk d O;kl nh?kZo`Ùk

2 2

2 2

x y

E : 1a b+ = (a > b) dh ukfHk;ksa dks feykus okyh js [kk rFkk

nwljk O;kl E dk y?kqv{k gS rks -

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Q U E S T I O N

B A

N K

2 0 1 4

(A) E dh mRds Unzrk 1

2 gksxhA

•(B) E dh mRds Unzrk 1

2

gksxhA

(C) ;fn E,2 2

2

x y1

4 b+ = gks rks o`Ùk C dk fu;ked o`Ùk 2 2x y 8+ = gksxkA

•(D) ;fn E,2 2

2

x y1

4 b+ = gks rks o`Ùk C dk fu;ked o`Ùk 2 2x y 4+ = gksxkA

9. Ans. (B,D)

b

ae

b = ae

b

2

= a

2

(1 – e

2

)Þ 2e2 = 1

1e

2=

if a = 2

2b 2

2= =

\ Director circle of C is x2 + y2 = 2 × 2 = 4

9. Consider an ellipse E1, whose major axis coincides with minor axis of ellipse ( )

2 2

2 2

x y

E : 1 a ba b+ = >and minor axis of E1 is segment joining foci of E. If e is eccentricity of E then-

•(A) eccentricity of E1 is

2

2

1 2e

1 e

--

(B) foci of E1 are ( )20,b 1 2e- and ( )20, b 1 2e- -

•(C) foci of E1 are ( )20,a 1 2e- and ( )20, a 1 2e- -

(D) If eccentricity of E1 is e1, then (1 – e12

) (1 – e

2

) = e12

ekuk d nh?kZo`Ùk E1, ftldk nh?kZ v{k nh?kZo`Ùk ( )2 2

2 2

x yE : 1 a b

a b+ = > ds y?kq v{k ds lEikrh gS rFkk E1 dk y?kq

v{k E dh ukfHk;ks a dks feykus okyk k.M gSA fn e, E dh mRds Uærk gks rks -

(A) E1 dh mRds Uærk

2

2

1 2e

1 e

--

gksxhA

(B) E1 dh ukfHk;k¡ ( )20, b 1 2e- rFkk ( )20, b 1 2e- - gks axhA

(C) E1 dh ukfHk;k¡ ( )20,a 1 2e- rFkk ( )

20, a 1 2e- - gks axhA

(D) ;fn E1 dh mRds Uærk e1 gks rks (1 – e12) (1 – e2) = e1

2 gksxkA

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Q U E S T I O N

B A

N K

2 0 1 4

9. Ans. (A,C)

2 2

2 2

x yE : 1

a b+ =

B

S' S'0

B

e2

=

2

2

b

1 a-2 2

1 2 2 2

x yE : 1

a e b+ =

b > aeeccentricity of E

1 S' 0

B'(0,–b)

B(0,b)

S(ae,0)(–ae,0)

2 2

1 2

a ee 1

b= -

2

22 21 1 2e1 e 1 e 1 e-ì ü= - =í ý- -î þ

foci of E1 are (0, ± be1)

i.e.

2

2

b. 1 2e0,

1 e

æ ö-±ç ÷

ç ÷-è ø

Þ ( )2

2b. 1 2e0, 0, a 1 2eb / a

æ ö-± Þ ± -ç ÷

ç ÷è ø

Paragraph for Question 11 & 12

A boy moving along the ellipse2 2x y

132 18

+ = , at each and every point on it he is drawing a tangent

and finding the area of triangle formed by it with co-ordinates axes. He found that area of triangleis > m and is m at P,Q,R and S.

,d ckyd nh?kZo`Ùk 2 2x y

132 18

+ = ds vuqfn'k xfr djrs gq bl nh?kZo`Ùk ds izR;s d fcUnq ij Li'kZ js [kk khapdj

funs Z'kh v{kkas kjk fufeZr f=Hkqt ds ks=Qy dks Kkr djrk gSA og ikrk gS fd f=Hkqt dk ks=Qy lnS o > m gS rFkk

fcUnqvks a P,Q,R rFkk S ij m gSA11. The value of 'm' is -

•(A) 24 (B) 24 (C) 24p (D) 24p'm' dk eku gksxk -

•(A) 24 (B) 24 (C) 24p (D) 24p12. Area of the quadrilateral PQRS is-

(A) 24 •(B) 48 (C) 2 24 (D) 24 prqHkqZt PQRS dk ks=Qy gksxk -

(A) 24 •(B) 48 (C) 2 24 (D) 24

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Q U E S T I O N

B A

N K

2 0 1 4

Paragraph for Question 11 & 12

11. Ans. (A)

CA

B0

Let A be ( )4 2 cos ,3 2 sinq q

equation of tangentx cos ysin

14 2 3 2

q q+ =

OB 4 2 sec= q , OC 3 2cosec= q

1 24Area 24sec cosec

2 sin 2

= q q =q

minimum area = 2412. Ans. (B)

For minimum area sin2q = 1

3 5 7, , ,

4 4 4 4

p p p pq =

Q P(4,3)

SR

area = 8 × 6 = 48

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